Question

Two blocks of masses of 40 kg and 30 kg are connected by a weightless string passing over a frictionless pulley as shown in the figure :

The acceleration of the system would be :

• A. $0.7 \mathrm {~ms} ^ { - 2 }$
• B. $0.8 \mathrm {~ms} ^ { - 2 }$
• C. $0.6 \mathrm {~ms} ^ { - 2 }$
• D. $0.5 \mathrm {~ms} ^ { - 2 }$

Answer 1

Answer: (A)

$0.7 \mathrm {~ms} ^ { - 2 }$

Answer 2

Here $m _ { 1 } = 40 \mathrm {~kg}$

$m _ { 2 } = 30 \mathrm {~kg}$

$\theta = 30 ^ { \circ }$

Let T be the tension in the sting and a be the acceleration of the system.

Their equation of motion are

$m _ { 1 } g \sin 30 ^ { \circ } - T = m _ { 1 } a$ ….. (i)

$T - m _ { 2 } g \sin 30 = m _ { 2 } a$ ….. (ii)

Adding (i) and (ii) we get

$\left( m _ { 1 } + m _ { 2 } \right) a = \left( m _ { 1 } - m _ { 2 } \right) g \sin 30 ^ { \circ }$

Substituting the given values we get

$( 40 + 30 ) a = ( 40 - 30 ) \times 9.8 \times \frac { 1 } { 2 } = 49$

$a = \frac { 49 } { 70 } = 0.7 \mathrm {~ms} ^ { - 2 }$